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2 years ago

Which of the following is an example of a nonmetal that is solid at room

Chemistry

1 answer:

Darya [45]2 years ago

6 0

Answer:

Carbon

Explanation:

At room temperature20–22 °C (68–72 °F), nitrogen and oxygen are gases, while bromine is a liquid.

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How many moles are in 2.5L of 1.75 M Na2CO3

mixas84 [53]

Answer: There are 4.375 moles in 2.5 L of 1.75 M $Na_2CO_3$

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of solution = 1.75 M

Volume of solution = 2.5 L

Putting values in equation , we get:

$1.75M=\frac{\text{Moles of} Na_2CO_3}{2.5L}\\\\\text{Moles of }Na_2CO_3=1.75mol/L\times 2.5L=4.375mol$

6 0

2 years ago

C time

lesya692 [45]

Answer:

B. Spring balance - a device used for measuring the weight or force of gravity acting on an object.

Explanation:

A Force is any interaction that changes the motion or position of an obkpjectbthatbit is interacting with. Whenever there is an interaction between two objects, there is a force exerted by each of the objects on one themselves.

Forces are generally divided into contact forces and non-contact over field forces.

In contact forces, the two objects physically in contact with each other. Examples of contact forces are push or pull forces, frictional forces, tensional forces, spring forces, etc.

Non-contact forces are forces in which the two objects interacting do no need to be physically in contact with one another. Examples include, gravitational forces, magnetic forces, electrical forces, etc.

Instruments used in measuring forces are known as force gauges.

From the instruments listed above:

A. A ruler is an instrument used in measuring length

B. Spring balance is a device used for measuring the weight or force of gravity acting on an object.

C. A thermometer is an instrument used in measuring temperature

D. A windbvane is an instrument used in measuring wind direction.

4 0

3 years ago

The reaction for photosynthesis producing glucose sugar and oxygen gas is:

Anvisha [2.4K]

1.5 grams of glucose is produced from 2.20 g of CO₂.

To find the mass of glucose produced, first you must know the balanced reaction. For this, the Law of Conservation of Matter is followed.

The law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.

So, in this case, the balanced reaction is:

6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the amounts of moles of each reactant and product participate in the reaction:

CO₂: 6 moles
H₂O: 6 moles
C₆H₁₂O₆: 1 mole
O₂: 6 moles

So, you know that 2.20 g of CO₂ react, whose molar weight is 44.01 g/mole. By definition of molar mass, 1 mole of CO₂ has 44.01 g. So, the number of moles that 2.20 grams of the compound represent is calculated as:

$moles of CO_{2} =2.20 grams*\frac{1 mole}{44.01 grams}$

moles of CO₂= 0.05 moles

Now you must follow the following rule of three: if by stoichiometry of the reaction 6 moles of CO₂ produce 1 mole of C₆H₁₂O₆, 0.05 moles of CO₂ produce how many moles of C₆H₁₂O₆?

$moles of C_{6} H_{12} O_{6} =\frac{0.05moles of CO_{2} *1 mole of C_{6} H_{12} O_{6}}{6moles of CO_{2}}$

moles of C₆H₁₂O₆= 8.33*10⁻³

Being the molar mass of glucose 180.18 g/mole, the mass that 8.33*10⁻³ moles of the compound represent is calculated as:

$mass of glucose =8.33*10^{-3} moles*\frac{180.18 grams}{1 mole}$

mass of glucose= 1.5 grams

Then, 1.5 grams of glucose is produced from 2.20 g of CO₂.

5 0

3 years ago

In a homogeneous mixture, the liquid is the?

scoray [572]

the solvent, hope this helps

5 0

3 years ago

Read 2 more answers

The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W

Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹

Explanation:

Assume the rate law is

rate = k[A][B]²

If you are comparing two rates,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}$

You are cutting each concentration in half, so

$\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}$

Then,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}$

8 0

2 years ago